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3 years ago

Gggggggggggggghhhhhhhhhhhhhiiiiiiiiii

Mathematics

2 answers:

sasho [114]3 years ago

4 0

That answer is 17 :):):):):)

likoan [24]3 years ago

3 0

Answer:

Minimum: 6

Lower Quartile (Q1): 9

Median (Q2): 15

Upper Quartile (Q3): 26

Maximum: 34

Interquartile Range: 17

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Answer:

21minutes

Step-by-step explanation:

its 21 minutes because 5x3=15 so 7x3=21

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3 years ago

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Dmitry_Shevchenko [17]

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The diameter of a circle is 6 what is the circumfrence

MA_775_DIABLO [31]

Answer:

18.84

Step-by-step explanation:

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3 years ago

Can you guys pls help me with this math question

Sindrei [870]

Answer:

Dimensions: 150 m x 150 m

Area: 22,500m²

Step-by-step explanation:

Given information:

Rectangular field
Total amount of fencing = 600m
All 4 sides of the field need to be fenced

Let $x$ = width of the field

Let $y$ = length of the field

Create two equations from the given information:

Area of field: $A= xy$

Perimeter of fence: $2(x + y) = 600$

Rearrange the equation for the perimeter of the fence to make y the subject:

$\begin{aligned} \implies 2(x + y) & = 600\\ x+y & = 300\\y & = 300-x\end{aligned}$

Substitute this into the equation for Area:

$\begin{aligned}\implies A & = xy\\& = x(300-x)\\& = 300x-x^2 \end{aligned}$

To find the value of x that will make the area a maximum, differentiate A with respect to x:

$\begin{aligned}A & =300x-x^2\\\implies \dfrac{dA}{dx}& =300-2x\end{aligned}$

Set it to zero and solve for x:

$\begin{aligned}\dfrac{dA}{dx} & =0\\ \implies 300-2x & =0 \\ x & = 150 \end{aligned}$

Substitute the found value of x into the original equation for the perimeter and solve for y:

$\begin{aligned}2(x + y) & = 600\\\implies 2(150)+2y & = 600\\2y & = 300\\y & = 150\end{aligned}$

Therefore, the dimensions that will give Tanya the maximum area are:

150 m x 150 m

The maximum area is:

$\begin{aligned}\implies \sf Area_{max} & = xy\\& = 150 \cdot 150\\& = 22500\: \sf m^2 \end{aligned}$

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2 years ago

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Leviafan [203]

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3 years ago